Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 4}{x - 9} = \dfrac{6x + 23}{x - 9}$
Multiply both sides by $x - 9$ $ \dfrac{x^2 - 4}{x - 9} (x - 9) = \dfrac{6x + 23}{x - 9} (x - 9)$ $ x^2 - 4 = 6x + 23$ Subtract $6x + 23$ from both sides: $ x^2 - 4 - (6x + 23) = 6x + 23 - (6x + 23)$ $ x^2 - 4 - 6x - 23 = 0$ $ x^2 - 27 - 6x = 0$ Factor the expression: $ (x - 9)(x + 3) = 0$ Therefore $x = 9$ or $x = -3$ At $x = 9$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 9$, it is an extraneous solution.